Molecules of benzoic acid (C$_{6}H_{5}$COOH) dimerise in benzene. $'w' \,g$ of the acid dissolved in $30\, g$ of benzene shows a depression in freezing point equal to $2K$. If the percentage association of the acid to form dimer in the solution is $80$, then $w$ is :
(Given that $K_{f} = 5 K\, kg\, mol^{-1} $, Molar mass of benzoic acid $= 122 \,g\, mol^{-1})$
- 1.8 g
- 2.4 g
- 1.0 g
- 1.5 g
The Correct Option is B
Solution and Explanation
To determine the weight \(w\) of benzoic acid required, we start by analyzing the freezing point depression data provided. Given that the molecules dimerize, we need to account for this in our calculations:
- Understanding Dimerization: When benzoic acid dimerizes, two molecules join to form one dimeric molecule. This affects the colligative properties, such as the depression in freezing point.
- Relating Freezing Point Depression to Molality:
- The formula for depression in freezing point is given by: \(\Delta T_f = i \cdot K_f \cdot m\) where:
- \(\Delta T_f = 2\, K\) (given)
- \(K_f = 5\, K\, kg\, mol^{-1}\) (given)
- \(m\) is the molality of the solution.
- \(i\) is the van't Hoff factor which accounts for the extent of association.
- The formula for depression in freezing point is given by: \(\Delta T_f = i \cdot K_f \cdot m\) where:
- Calculating the Van't Hoff Factor:
- For dimerization, the van't Hoff factor \(i\) is given by: \(i = 1 - \alpha/2\) where \(\alpha\) is the degree of association.
- Given \(\alpha = 0.8\) (80%), we have: \(i = 1 - 0.8/2 = 0.6\).
- Substitute values into the freezing point depression formula:
- \(2 = 0.6 \times 5 \times m\)
- Solve for molality \(m\): \(m = \frac{2}{0.6 \times 5} = \frac{2}{3} = 0.6667\, mol/kg\).
- Calculate the weight of benzoic acid:
- Molality (\(m\)) is defined as moles of solute per kg of solvent. So, \(m = \frac{w/M}{0.03}\) (since 30 g of benzene is used, equivalent to 0.03 kg) where \(M\) is the molar mass of benzoic acid (122 g/mol).
- Rearranging gives: \(\frac{w}{122 \times 0.03} = 0.6667\)
- Solve for \(w\): \(w = 122 \times 0.03 \times 0.6667 = 2.439\, g \approx 2.4\, g\) (rounded to 2.4 g for convenience).
Thus, the correct answer for the weight of benzoic acid that results in the observed depression in freezing point is 2.4 g.
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.